The length of the normal from ongin to the plane `x+2y-2z=9` is equal to :

Options

A. 2 units
B. 3 units
C. 4 units
D. 5 units

Answer

To find the length of the normal from the origin to the plane given by the equation \(x + 2y - 2z = 9\), we can use the formula for the distance from a point to a plane.

### Step-by-Step Solution:

1. **Identify the Plane Equation**:
The equation of the plane is given as:
\[ x + 2y - 2z - 9 = 0 \]
Here, we can identify \(A = 1\), \(B = 2\), \(C = -2\), and \(D = -9\).

2. **Identify the Point**:
The point from which we want to find the distance is the origin, which has coordinates \((x_1, y_1, z_1) = (0, 0, 0)\).

3. **Substitute into the Distance Formula**:
The formula for the distance \(D\) from a point \((x_1, y_1, z_1)\) to the plane \(Ax + By + Cz + D = 0\) is given by:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting the values:
\[ D = \frac{|1(0) + 2(0) - 2(0) - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}} \]

4. **Calculate the Numerator**:
The numerator simplifies to:
\[ |0 + 0 + 0 - 9| = |-9| = 9 \]

5. **Calculate the Denominator**:
The denominator simplifies to:
\[ \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

6. **Calculate the Distance**:
Now substituting back into the distance formula:
\[ D = \frac{9}{3} = 3 \]

### Conclusion:
The length of the normal from the origin to the plane \(x + 2y - 2z = 9\) is \(3\) units.